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A standard exercise in essentially any introductory textbook on general relativity is to work out the non-relativistic limit of the 3+1D Einstein field equations. This is most commonly done in order to nail down the proportionality constant between the Einstein tensor $G$ and stress-energy tensor $T$ that reproduces Newton's law of universal gravitation in the nonrelativistic regime.

That is, one typically starts with the ansatz $G = \kappa T$, where $\kappa$ is an unknown constant, and then works out that in the nonrelativistic limit, this reduces to $$\nabla^2 \Phi = \frac{1}{2} \kappa \rho, \tag{1}$$ which reproduces Newton's law of gravitation if we set $\kappa = 8 \pi G$.

What happens if we take the same nonrelativistic limit in arbitrary dimensions?

My guess is that we get the same limiting expression (1) in $D \geq 4$ spacetime dimensions, but we get something qualitatively different for $D = 2$ or $D = 3$. That's because in those lower dimensions, the Weyl tensor vanishes, so I don't think you can have long-distance gravitational effects that extend through vacuum. But I have no idea what you do get.

This seems like the kind of thing that someone would have already worked out somewhere, but I couldn't find it.

tparker
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  • Using classical physics, in $n$ spatial dimensions one can deduce that Newtonian force of gravity has the form $F\propto \frac{1}{r^{n-1}}$ So I assume that since Einsteinian gravity reduces to the force law $\frac{1}{r^2}$ in the Newtonian limit, it should reduce to the same ($\frac{1}{r^{n-1}}$) for arbitrary dimensions. – joseph h Jan 18 '23 at 05:05
  • @josephh That's a very natural assumption, but unfortunately it's incorrect. As John Baez says in a comment at https://www.math.columbia.edu/~woit/wordpress/?p=555, $(2+1)D$ Einstein gravity does not reduce to the Newtonian expression in the nonrelativistic limit. Hence my question. – tparker Jan 18 '23 at 05:28
  • But aren't you interested $D\ge 4$? Or are you asking specifically about 2+1? – joseph h Jan 18 '23 at 05:34
  • @josephh As I said in both the title and the body of the question: arbitrary dimensions, so both greater than and less than 4. – tparker Jan 18 '23 at 05:52
  • Do you want the number of time dimensions to be arbitrary, or $1$? – J.G. Jan 18 '23 at 07:34
  • Related: https://physics.stackexchange.com/q/200020/2451 , https://physics.stackexchange.com/q/211930/2451 and links therein. – Qmechanic Jan 18 '23 at 10:28
  • @J.G. One time dimension. With multiple time dimensions, the low-speed and low-curvature limit would approach something very different from Newtonian physics, so I wouldn’t really call that “the nonrelativistic limit”. I would have called it “the Newtonian limit” in my question, but that would run the risk of ambiguity with Newtonian gravitation, while the whole point of my question is that we don’t get standard Newtonian gravitation in that limit. – tparker Jan 18 '23 at 13:14
  • @tparker Have you looked into https://physics.stackexchange.com/q/75473 (Behavior of black holes in higher- and lower-dimensional space-times) and maybe here https://arxiv.org/abs/2005.06809v1 (Maximum Force and Naked Singularities in Higher Dimensions)? – JanG Jan 18 '23 at 17:01

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